University of Alberta Math 14 Sample Exam Math 14 Solutions 1. Test the following series for convergence or divergence (a) ( n +n+1 3n +n+1 )n, (b) 3 n (n +1) (c) SOL: n!, arccos( n n +1 ), (a) ( n +n+1 3n +n+1 )n, We use the root test: n ( n + n + 1 3n + n + 1 )n = n + n + 1 3n + n + 1 = 3 < 1. This series is CONVERGENT. (b) 3 n (n +1), n! We use the ratio test: 3 n+1 ((n+1) +1) (n+1)! 3 n (n +1) n! This series is CONVERGENT. 3 n+1 ((n + 1) + 1) n! = (n + 1)! 3 n (n + 1) = 3((n + 1) + 1) (n + 1)(n + 1) = 0 < 1. (c) n arccos( ), n +1 Since arccos( n n + 1 ) = arccos(0) = π 0. this series is DIVERGENT by the divergence test.. Find the radius of convergence and the interval of convergence for n+1 (x+3) n+1 (n+1) n+1+1 n (x+3) n n n+1 n (x + 3) n n n + 1. = (x + 3)(n n + 1) (n + 1) n + 1 + 1 = (x + 3).
So the series is convergent when x + 3 < 1/ and divergent when x + 3 > 1/. When x + 3 = 1 we have 1 1 n 3. n n+1 When x + 3 = 1 we have ( 1) n series test. n n+1. which is convergent by comparison to. which is convergent by Alternating Thus the Radius of Convergence is 1 and the interval of convergence is [ 7, 5]. 3. Find the Taylor series for f(x) = x at x = 1. f (n) (x) = 1 1 3 The Taylor series is f(x) = x 1 f (x) = 1 x 1 f (x) = 1 1 f (x) = 1 1 x 3 3 x 5 3)... (n x (n+1) n 1 1 3 5... (n 3) = ( 1) x (n+1) n. n 1 1 3 5... (n 3) ( 1) (x 1) n. n n! n=0 4. (a) Find the slope of the tangent line to the curve r = 1 + cos(θ) at the point where θ = π. (b) Find the area of the region that lies inside the curve r = 1 + cos(θ) and outside the circle r = 1. dy dx = cos(θ) sin (θ) + cos (θ) sin(theta) cos(θ) sin(θ). At θ = π dy we get = 1. dx (b) The two curves intersect when cos(θ) = 0 which is when θ = ± π. A = 1 π π (1 + cos(θ)) 1dθ = 1 π π cos(θ) + cos (θ)dθ
= + 1 4 π π 1 + cos(θ)dθ = + π 4. 5. Let P 1 (1,, 1), P (, 3, ), P 3 (3, 7, 1). (a) Find an equation for the plane passing through P 1, P and P 3. (b) Find Find the area of the triangle with the vertices P 1, P and P 3. P 1 P = (1, 1, 3) P 1 P 3 = (, 5, ) P 1 P P 1 P 3 = (13, 4, 3). (a) 13(x 1) 4(y ) + 3(z 1) = 0 or 13x 4y + 3z = 8. (b) Area = 1 P 1 P P 1 P 3 = 169+16+9 = 194. 6. Find the projection of u =< 1, 3, 5 > in the direction of v =< 3,, 4 >. proj v (u) = u v v v v = 9 v =< 3,, 4 >. 9 7. Let r(t) =< 4 sin(t), 3t, 4 cos(t) >. (a) Find T, N, B. (b) Find the curvature at t = π. (a) r =< 4 cos(t), 3, 4 sin(t) >. T =< 4 5 cos(t), 3 5, 4 5 sin(t) >. T =< 4 5 sin(t), 0, 4 5 cos(t) >. N =< sin(t), 0, cos(t) >. 3
B = T N =< 3 5 cos(t), 4 5, 3 5 sin(t) >. (b) K = T 4 r = 5 5 = 4 5. 8. Find the it or show that the it does not exist: xy 1 (x,y) (1,1) x + y x y +. Consider the lines y = m(x 1) + 1 passing through (1, 1). When we approach (1, 1) along these lines we get: x 1 xm(x 1) + x 1 x + (m(x 1) + 1) x (m(x 1) + 1) + = x 1 (x 1)(mx + 1) (x 1)[x 1 + m(x 1) + (m 1) m] = mx + 1 = x 1 x 1 + m(x 1) + (m 1) m = m + 1 Since the it depends on m, we get two different answers for two different lines, thus the it doesn t exist. Note: You can actually calculate just the it on the line y = 1 and y = x and get two different answers in this case. Also, if you have troubles factoring (x 1) you should just use L Hospital. 9. Let f(x, y, z) = x + 3xy y + z 1. (a) Find f at P (1, 1, 1). (b) Find the directional derivative of f in the direction of the vector u = (, 1, ). (a) f =< x + 3y, 3x 4y, z >. At P (1, 1, 1 we have f =< 5, 1, >. (b) The direction of u is u u =< 3, 1 3, 3 >. Hence D u (f) =< 5, 1, > < 3, 1 3, 3 >= 15 3 = 5. 4
10. Find equations for the tangent plane and normal line to the surface x y + 4z 3 = 3 at the point P (1, 1, 1). At P (1, 1, 1) we have g =< 1, 4, 1 >. Thus the tangent plane is or (x 1) 4(y 1) + 1(z 1) = 0, x 4y + 1z = 9. The normal line is x y z = 1 + t = 1 4t = 1 + 1t 11. Let z = x 3 + x y and let x = s + t, y = st. Use the chain rule to calculate z z and. s t z s = z x x s + z y y s = (3x +4xy)s+x t = [3(s +t) +4(s +t)st]s+(s +t) t, z t = z x x t + z y y t = (3x +4xy)+x s = [3(s +t) +4(s +t)st]+(s +t) s. 1. Find and classify the critical points of the function f(x, y) = x 3 y 3 3x + 3y +. { fx = 3x 3 = 0 f y = 3y + 3 = 0 Thus x = ±1 and y = ±1. The Hessian is H = 6x 0 0 6y = 36xy 5
(1,1). H = 36 < 0 thus (1, 1) is a saddle point. (-1,-1). H = 36 < 0 thus ( 1, 1) is a saddle point. (-1,1). H = 36 > 0 and f xx = 6 < 0 thus ( 1, 1) is a local maximum. (1,-1). H = 36 > 0 and f xx = 6 > 0 thus (1, 1) is a local minimum. 13. Find the absolute minimum and maximum of f(x, y) = x + xy y 5y on the rectangle 0 x 3, 4 y. Critical points inside: The only CP is (1, ). On the Boundary { fx = x + y = 0 f y = x y 5 = 0 x = 0. Then f(y) = y 5y which has only CP y = 5. We get (0, 4), (0, 5 ), (0, ). x = 3. Then f(y) = 9 y y which has only CP y = 1. We get (3, 4), (3, 1), (3, ). y = 4. Then f(x) = x 4x + 4 which has only CP x =. We get (0, 4), (, 4), (3, 4). y =. Then f(x) = x + x 14 which has only CP x = 1. We get (0, ), ( 1, ), (3, ). f(0, 4) = 4 f(0, 5) = 5 4 f(0, ) = 14 f(3, 4) = 1 f(3, 1) = 10 absolute max f(3, ) = 1 f(, 4) = 0 f( 1, ) = 15 absolute min 14. Find Find the minimum and maximum of f(x, y) = xy + 14 subject to x + y = 18. f = λ g (y, x) = λ(x, y) 6
y = λx x = λy x + y = 18 λ = y x = x y x = y x = y. From the last equation we get x = y = 9. Hence x = ±3, y = ±3. Since x + y = 18 is a circle, which is a bounded closed region in plane, f attains its max/min. f(3, 3) = 3 absolute max f(3, 3) = 5 absolute min f( 3, 3) = 5 absolute min f( 3, 3) = 3 absolute max 7